Two or more surds can be compared without actually calculating their values, depending on the types of surds. Sometimes, it is possible to compare them without solving them in detail. Here are some of these cases:

• The two surds are \(\sqrt a  + \sqrt b \) and \(\sqrt c  + \sqrt d \)  where \(a + b = c + d\), then 

  \(\sqrt a  + \sqrt b  > \sqrt c  + \sqrt d \)  if  \(\left| {a - b} \right| < \left| {c - d} \right|\), for example

  \(\sqrt {13}  + \sqrt {17}  > \sqrt {12}  + \sqrt {18} \) as \(13 + 17 = 12 + 18\), but the difference between the numbers \(13\) and \(17\) is less than the difference between \(12\) and \(18\). Lesser is the difference between the numbers, and more is the sum of their square roots, given that the sum of the numbers is constant.




• The two surds are \(\sqrt a  - \sqrt b \) and \(\sqrt {a + k}  - \sqrt {b + k} \), where \(k\) is a positive integer, then \(\sqrt a  - \sqrt b  > \sqrt {a + k}  - \sqrt {b + k} \).

  For example \(\sqrt a  - \sqrt b  > \sqrt {a + 1}  - \sqrt {b + 1}  > \sqrt {a + 2}  - \sqrt {b + 2} \)

  In general for any fractional power, \({\left( a \right)^n} - {\left( b \right)^n} > {\left( {a + k} \right)^n} - {\left( {b + k} \right)^n}\), where \(0 < n < 1\).

  For example \(\sqrt {1000}  - \sqrt {999}  < \sqrt {10}  - \sqrt 9 \), \(\sqrt[3]{{50}} - \sqrt[3]{{49}} < \sqrt[3]{{40}} - \sqrt[3]{{39}}\) etc.  




• In some cases, a direct comparison is not possible, in that case, we can take square both sides and then compare them. For example, suppose \(a = \sqrt {35}  + \sqrt {11} \) and \(b = \sqrt {55}  + \sqrt 7 \), squaring both the surds, we have,

  \({a^2} = 35 + 11 + 2\sqrt {385} \)

  \({b^2} = 55 + 7 + 2\sqrt {385} \)

  Clearly \({b^2}\) is bigger or \(b > a\). 

Example 01: Arrange the following simple surds in descending order
\(\sqrt 8 ,\,\,\sqrt[3]{{24}},\;\sqrt[6]{{500}}\)

Example 02: Arrange the following numbers in ascending order
\(a = 3 + \sqrt 2 ,\;b = \sqrt 5  + \sqrt 7 ,\;c = \sqrt {24} \)